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Since $\zeta(s)=0$ and $\Gamma(s)$ has poles at even negative numbers $-2, -4, -6...$ Can we find a value for $\frac{1}{\zeta(s)\Gamma(s)}$ using L'Hôpital's rule?

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    $\begingroup$ The command of Mathematica Limit[1/Zeta[s]/Gamma[s], s -> -2*n, Assumptions -> n \[Element] Integers && n > 0] produces $$ \frac{(2 n)!}{\zeta '(-2 n)}.$$ $\endgroup$
    – user64494
    Commented Jul 26, 2020 at 5:51
  • $\begingroup$ @user64494 Thank you. $\endgroup$
    – user155294
    Commented Jul 26, 2020 at 6:26
  • $\begingroup$ The functional equation for $\zeta(s)$ gives a formula for $\zeta(n)$, $n=0,-1,-2,\cdots$ in terms of Bernoulli numbers. As you mentioned, this formula just gives 0 for negative even integers. The value you are interested in is equivalent to computing $\zeta(3), \zeta(5), \cdots$, but these are much more mysterious, with no closed formulae. In fact, despite there being a closed formula for even natural numbers, just proving that $\zeta(3)$ is irrational ended up a very subtle argument by Apery, and there are few irrationality results for the others. $\endgroup$
    – Ralph Furman
    Commented Jul 27, 2020 at 19:01

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